We know that Normally, the number of voltage intervals is given by

a). Resolution

• $N=2^M \cdots\mathrm{Equation \ 1}$.......(1)

where M is the ADC's resolution in bits

• Thus from equation 1 , we obtain

$Resolution \space (N)=2^8=\bf256 \space \text{quantization levels}$

• ADC voltage resolution is given by,

$Q = \dfrac{V_i }{ (2^M-1)} mV/LSB \cdots \mathrm{Equation \ 1}$

and from equation 2, we have

$Q = \dfrac{5.1}{ (2^8-1)} mV/LSB=\bf20 mV/LSB$

$\therefore$ We can say that to change output by $1 \ \mathrm {LSB}$ we have to change input by $20 \space mV$  

2. Digital Output

• For$1.28$V analog input, digital output can be calculated as,

$D= \dfrac{1.28 V}{20\ mV/LSB}=\bf64\ LSBs$

The binary equivalent of $64 $ is $0100\space 0000_2$.